Contiguity of measures

Remarks

Contiguity is the asymptotic analogue of absolute continuity; for constant sequences $P_n = P$ and $Q_n = Q$ it is the same as $Q \ll P$. If $Q_n \triangleleft P_n$, then every sequence of statistics that tends to $0$ in $P_n$-probability also tends to $0$ in $Q_n$-probability.

Example: let $P_n = \mathcal{N}(0, 1)^{\otimes n}$ and $Q_n = \mathcal{N}(h/\sqrt{n}, 1)^{\otimes n}$, products of normal distributions, for a fixed $h \in \R$. Under $P_n$, the log-likelihood ratio $\log(dQ_n/dP_n) = h n^{-1/2}\sum_i X_i - h^2/2$ has the law $\mathcal{N}(-h^2/2, h^2)$ for every $n$. By Le Cam's first lemma, a limit law $L$ of $\log(dQ_n/dP_n)$ under $P_n$ with $E e^{L} = 1$ implies $Q_n \triangleleft P_n$, and here $E e^{L} = e^{-h^2/2 + h^2/2} = 1$; by symmetry the sequences are mutually contiguous.

Non-example: let $P_n = \mathcal{N}(0, 1)^{\otimes n}$ and $Q_n = \mathcal{N}(1, 1)^{\otimes n}$, and let $A_n = \set{\bar{x}_n > 1/2}$, with $\bar{x}_n$ the mean of the coordinates. Then $P_n(A_n) = P(Z > \sqrt{n}/2) \to 0$ but $Q_n(A_n) = P(Z > -\sqrt{n}/2) \to 1$ for $Z$ standard normal, so $Q_n$ is not contiguous with respect to $P_n$.